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Question-163481

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Question Number 163481 by smallEinstein last updated on 07/Jan/22
Answered by Mathspace last updated on 07/Jan/22
Ψ_n =∫_0 ^∞ cos(x^n )logx dx letf(a)=∫_0 ^∞ x^a cos(x^n )dx f^′ (a)=∫_0 ^∞ x^a logx cos(x^n )dx ⇒f^′ (0)=∫_0 ^∞ cos(x^n )logx dx f(a)=Re(∫_0 ^∞ x^a e^(ix^n ) dx) ∫_0 ^∞ x^a e^(ix^n ) dx let ix^n =−t⇒x^n =it ⇒ x=(it)^(1/n) =e^((iπ)/(2n)) t^(1/(n )) ⇒ ∫_0 ^∞ x^a e^(ix^n ) dx=e^((iπ)/(2n)) ∫_0 ^∞ e^((iπa)/(2n)) t^(a/n) (1/n)t^((1/n)−1) e^(−t) dt =(1/n)e^(((iπ)/(2n))(a+1)) ∫_0 ^∞ t^(((a+1)/n)−1) e^(−t) dt =(1/n)Γ(((a+1)/n))e^(((iπ)/(2n))(a+1)) ⇒f(a)=(1/n)Γ(((a+1)/n))cos(((π(a+1))/(2n))) ⇒f^′ (a)=(1/n^2 )Γ^′ (((a+1)/n))cos(((π(a+1))/(2n))) −(π/(2n^2 ))sin(((π(a+1))/(2n)))Γ(((a+1)/n)) Ψ_n =f^′ (0)=(1/n^2 )Γ^′ ((1/n))cos((π/(2n))) −(π/(2n^2 ))sin((π/(2n)))Γ((1/n)) Ψ_4 =(1/(16))Γ^′ ((1/4))cos((π/8))−(π/(32))Γ((1/4))sin((π/8)) =∫_0 ^∞ cos(x^4 )logx dx
$$\Psi_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{{n}} \right){logx}\:{dx} \\ $$$${letf}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {cos}\left({x}^{{n}} \right){dx} \\ $$$${f}^{'} \left({a}\right)=\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {logx}\:{cos}\left({x}^{{n}} \right){dx} \\ $$$$\Rightarrow{f}^{'} \left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} {cos}\left({x}^{{n}} \right){logx}\:{dx} \\ $$$${f}\left({a}\right)={Re}\left(\int_{\mathrm{0}} ^{\infty} \:{x}^{{a}} {e}^{{ix}^{{n}} } {dx}\right)\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{x}^{{a}} \:{e}^{{ix}^{{n}} } {dx} \\ $$$${let}\:{ix}^{{n}} =−{t}\Rightarrow{x}^{{n}} ={it}\:\Rightarrow \\ $$$${x}=\left({it}\right)^{\frac{\mathrm{1}}{{n}}} ={e}^{\frac{{i}\pi}{\mathrm{2}{n}}} \:{t}^{\frac{\mathrm{1}}{{n}\:}} \Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{x}^{{a}} \:\:{e}^{{ix}^{{n}} } {dx}={e}^{\frac{{i}\pi}{\mathrm{2}{n}}} \int_{\mathrm{0}} ^{\infty} {e}^{\frac{{i}\pi{a}}{\mathrm{2}{n}}} \:{t}^{\frac{{a}}{{n}}} \:\frac{\mathrm{1}}{{n}}{t}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \:{e}^{−{t}} {dt} \\ $$$$=\frac{\mathrm{1}}{{n}}{e}^{\frac{{i}\pi}{\mathrm{2}{n}}\left({a}+\mathrm{1}\right)} \:\int_{\mathrm{0}} ^{\infty} \:{t}^{\frac{{a}+\mathrm{1}}{{n}}−\mathrm{1}} \:{e}^{−{t}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{{n}}\Gamma\left(\frac{{a}+\mathrm{1}}{{n}}\right){e}^{\frac{{i}\pi}{\mathrm{2}{n}}\left({a}+\mathrm{1}\right)} \\ $$$$\Rightarrow{f}\left({a}\right)=\frac{\mathrm{1}}{{n}}\Gamma\left(\frac{{a}+\mathrm{1}}{{n}}\right){cos}\left(\frac{\pi\left({a}+\mathrm{1}\right)}{\mathrm{2}{n}}\right) \\ $$$$\Rightarrow{f}^{'} \left({a}\right)=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\Gamma^{'} \left(\frac{{a}+\mathrm{1}}{{n}}\right){cos}\left(\frac{\pi\left({a}+\mathrm{1}\right)}{\mathrm{2}{n}}\right) \\ $$$$−\frac{\pi}{\mathrm{2}{n}^{\mathrm{2}} }{sin}\left(\frac{\pi\left({a}+\mathrm{1}\right)}{\mathrm{2}{n}}\right)\Gamma\left(\frac{{a}+\mathrm{1}}{{n}}\right) \\ $$$$\Psi_{{n}} ={f}^{'} \left(\mathrm{0}\right)=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\Gamma^{'} \left(\frac{\mathrm{1}}{{n}}\right){cos}\left(\frac{\pi}{\mathrm{2}{n}}\right) \\ $$$$−\frac{\pi}{\mathrm{2}{n}^{\mathrm{2}} }{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)\Gamma\left(\frac{\mathrm{1}}{{n}}\right) \\ $$$$\Psi_{\mathrm{4}} \:=\frac{\mathrm{1}}{\mathrm{16}}\Gamma^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right){cos}\left(\frac{\pi}{\mathrm{8}}\right)−\frac{\pi}{\mathrm{32}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right){sin}\left(\frac{\pi}{\mathrm{8}}\right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} {cos}\left({x}^{\mathrm{4}} \right){logx}\:{dx} \\ $$

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